\(a\text{) }\overrightarrow{AB}-\overrightarrow{CD}=\left(\overrightarrow{AC}+\overrightarrow{CB}\right)-\overrightarrow{CD}\\ =\overrightarrow{AC}-\left(\overrightarrow{CD}-\overrightarrow{CB}\right)=\overrightarrow{AC}-\overrightarrow{BD}\)

\(b\text{) }\overrightarrow{AB}+\overrightarrow{DC}+\overrightarrow{BD}+\overrightarrow{CA}=\left(\overrightarrow{AB}+\overrightarrow{BD}\right)+\left(\overrightarrow{DC}+\overrightarrow{CA}\right)\\ =\left(\overrightarrow{AB}+\overrightarrow{BD}\right)+\left(\overrightarrow{DC}+\overrightarrow{CA}\right)=\overrightarrow{AD}+\overrightarrow{DA}=0\)

\(c\text{) }\overrightarrow{AC}+\overrightarrow{DE}-\overrightarrow{DC}-\overrightarrow{CE}+\overrightarrow{CB}\\ =\left(\overrightarrow{AC}+\overrightarrow{CB}\right)+\left(\overrightarrow{DE}-\overrightarrow{DC}\right)-\overrightarrow{CE}\\ =\overrightarrow{AB}+\overrightarrow{CE}-\overrightarrow{CE}=\overrightarrow{AB}\)

\(d\text{) }\overrightarrow{AB}+\overrightarrow{DE}+\overrightarrow{CF}\\ =\left(\overrightarrow{AC}+\overrightarrow{CB}\right)+\left(\overrightarrow{DF}+\overrightarrow{FE}\right)+\left(\overrightarrow{CE}+\overrightarrow{EF}\right)\\ =\overrightarrow{AC}+\overrightarrow{CE}+\overrightarrow{CB}+\overrightarrow{DF}+\left(\overrightarrow{FE}+\overrightarrow{EF}\right)\\ =\overrightarrow{AC}+\overrightarrow{CE}+\overrightarrow{CB}+\overrightarrow{DF}\)

a)

\(\begin{array}{l}\overrightarrow {AB}  + \overrightarrow {CD}  = \overrightarrow {AD}  + \overrightarrow {CB} \\ \Leftrightarrow \overrightarrow {AB}  - \overrightarrow {CB}  = \overrightarrow {AD}  - \overrightarrow {CD} \\ \Leftrightarrow \overrightarrow {AB}  + \overrightarrow {BC}  = \overrightarrow {AD}  + \overrightarrow {DC} \\ \Leftrightarrow \overrightarrow {AC}  = \overrightarrow {AC} \end{array}\)

(luôn đúng)

b) \(\overrightarrow {AB}  + \overrightarrow {CD}  + \overrightarrow {BC}  + \overrightarrow {DA}  = \overrightarrow 0 \)

Ta có:

\(\begin{array}{l}\overrightarrow {AB}  + \overrightarrow {CD}  + \overrightarrow {BC}  + \overrightarrow {DA}  = (\overrightarrow {AB}  + \overrightarrow {BC} ) + (\overrightarrow {CD}  + \overrightarrow {DA} )\\ = \overrightarrow {AC}  + \overrightarrow {CA}  = \overrightarrow 0 \end{array}\)

Chú ý khi giải

+) Hiệu hai vecto chung gốc: \(\overrightarrow {AB}  - \overrightarrow {AC}  = \overrightarrow {CB} \) (suy ra từ tổng \(\overrightarrow {AB}  = \overrightarrow {AC}  + \overrightarrow {CB} \))

+) Với 4 điểm A, B, C, D bất kì ta có: \(\overrightarrow {AB}  + \overrightarrow {BC}  + \overrightarrow {CD}  + \overrightarrow {DA}  = \overrightarrow {AA}  = \overrightarrow 0 \)

a)

 \(\begin{array}{l}\overrightarrow {AB}  + \overrightarrow {BC}  + \overrightarrow {CD}  + \overrightarrow {DA}  = \left( {\overrightarrow {AB}  + \overrightarrow {BC} } \right) + \left( {\overrightarrow {CD}  + \overrightarrow {DA} } \right)\\ = \overrightarrow {AC}  + \overrightarrow {CA}  = \overrightarrow {AA}  = \overrightarrow 0 .\end{array}\)

b)

\(\overrightarrow {AC}  - \overrightarrow {AD}  = \overrightarrow {DC} \) và \(\overrightarrow {BC}  - \overrightarrow {BD}  = \overrightarrow {DC} \)

\( \Rightarrow \overrightarrow {AC}  - \overrightarrow {AD}  = \overrightarrow {BC}  - \overrightarrow {BD} \)

Ta có: \(\overrightarrow {AB}  + \overrightarrow {BC}  = \overrightarrow {AC}  \Leftrightarrow \overrightarrow {BC}  = \overrightarrow b  - \overrightarrow a \)

Lại có: vecto \(\overrightarrow {BD} ,\overrightarrow {BC} \) cùng hướng và \(\left| {\overrightarrow {BD} } \right| = \frac{1}{3}\left| {\overrightarrow {BC} } \right|\)

\( \Rightarrow \overrightarrow {BD}  = \frac{1}{3}\overrightarrow {BC}  = \frac{1}{3}(\overrightarrow b  - \overrightarrow a )\)

Tương tự: vecto \(\overrightarrow {BE} ,\overrightarrow {BC} \) cùng hướng và \(\left| {\overrightarrow {BE} } \right| = \frac{2}{3}\left| {\overrightarrow {BC} } \right|\)

\( \Rightarrow \overrightarrow {BE}  = \frac{2}{3}\overrightarrow {BC}  = \frac{2}{3}(\overrightarrow b  - \overrightarrow a )\)

Ta có:

\(\overrightarrow {AB}  + \overrightarrow {BD}  = \overrightarrow {AD}  \Leftrightarrow \overrightarrow {AD}  = \overrightarrow a  + \frac{1}{3}(\overrightarrow b  - \overrightarrow a ) = \frac{2}{3}\overrightarrow a  + \frac{1}{3}\overrightarrow b \)

\(\overrightarrow {AB}  + \overrightarrow {BE}  = \overrightarrow {AE}  \Leftrightarrow \overrightarrow {AE}  = \overrightarrow a  + \frac{2}{3}(\overrightarrow b  - \overrightarrow a ) = \frac{1}{3}\overrightarrow a  + \frac{2}{3}\overrightarrow b \)

a/ \(VT=\overrightarrow{AB}+\overrightarrow{BF}+\overrightarrow{BC}+\overrightarrow{CG}+\overrightarrow{CD}+\overrightarrow{DH}+\overrightarrow{DA}+\overrightarrow{AE}\)

\(=\left(\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA}\right)+\left(\frac{1}{2}\overrightarrow{BC}+\frac{1}{2}\overrightarrow{CD}+\frac{1}{2}\overrightarrow{DA}+\frac{1}{2}\overrightarrow{AB}\right)\)

\(=\overrightarrow{0}+\frac{1}{2}.\overrightarrow{0}=\overrightarrow{0}=VP\)

b/ Câu này áp dụng luôn kq câu a

\(\overrightarrow{MF}-\overrightarrow{MA}+\overrightarrow{MG}-\overrightarrow{MB}+\overrightarrow{MH}-\overrightarrow{MC}+\overrightarrow{ME}-\overrightarrow{MD}=\overrightarrow{0}\)

chuyển mấy cái vecto kia sang vế phải là có ngay đpcm câu b

c/\(VT=\overrightarrow{AI}+\overrightarrow{IB}+\overrightarrow{AI}+\overrightarrow{IC}+\overrightarrow{AI}+\overrightarrow{ID}=3\overrightarrow{AI}+\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{ID}\)

Để ý tới G là TĐ CD, F là TĐ BC

Theo quy tắc trung điểm

\(\Rightarrow\overrightarrow{IB}+\overrightarrow{IC}=2\overrightarrow{IF}=2\overrightarrow{HI}\)

\(\Rightarrow\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{ID}=2\overrightarrow{HI}+\overrightarrow{ID}=\overrightarrow{HI}+\overrightarrow{HD}\)

Mà \(\overrightarrow{HD}=\overrightarrow{AH}\Rightarrow\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{ID}=\overrightarrow{HI}+\overrightarrow{AH}=\overrightarrow{AI}\)

Thay vào cái trên sẽ có đpcm

a, =CD+FA+AB+DE+BC+EF=(CD+DE)+(AB+BC)+FA+EF

=CE+AC+FA+EF= (CE+EF)+AC+FA=CF+AC+FA=(CF+FA)+AC=CA+AC=0

b,VP=CD+AE+BF

VT=AD+FC+BE=AC+CD+CB+BF+BA+AE=(AC+CB)+CD+BF+BA+AE

=AB+CD+BF+BA+AE=(AB+BA)+CD+BF+AE=CD+BF+AE=VP(dccm)

\(\overrightarrow{AB}+\overrightarrow{CD}+\overrightarrow{BC}=\left(\overrightarrow{AB}+\overrightarrow{BC}\right)+\overrightarrow{CD}\)

\(=\overrightarrow{AC}+\overrightarrow{CD}=\overrightarrow{AD}\) (đpcm)